Answer
$\frac{-3\sqrt{2} + 3i\sqrt{6}}{\sqrt{6} + i\sqrt{2}}$
= $3i$
Work Step by Step
Now, $-3\sqrt{2} + 3i\sqrt{6} = -3\sqrt{2}(1 - \sqrt{3}i)$ where $(1 - \sqrt{3}i)$ is at $300^\circ$ with absolute value $\sqrt{1^2 + (-\sqrt{3})^2} = 2$, and $\sqrt{6} + i\sqrt{2} = \sqrt{2}(\sqrt{3} + i)$ where $(\sqrt{3} + i)$ is at $30^\circ$ with absolute value $\sqrt{(\sqrt{3})^2 + 1^2} = 2$
Therefore, $\frac{-3\sqrt{2} + 3i\sqrt{6}}{\sqrt{6} + i\sqrt{2}}$
= $\frac{-3\sqrt{2}(1 - \sqrt{3}i)}{\sqrt{2}(\sqrt{3} + i)}$
= $-3\cdot \frac{2cis300^\circ}{2cis30^\circ}$
= $-3cis(300^\circ - 30^\circ)$ (Quotient Theorem)
= $-3cis270^\circ$
= $-3(0 - i)$
= $3i$
